Advanced Calculus Single Variable

4.9.3 Shrinking Diameters

It is useful to consider another version of the nested interval lemma. This involves a sequence
of sets such that set

(n + 1)

is contained in set n and such that their diameters converge to 0.
It turns out that if the sets are also closed, then often there exists a unique point in all of
them.

Definition 4.9.20Let S be a nonempty set. Thendiam

(S)

is definedas

diam (S) ≡ sup{|x− y| : x,y ∈ S}.

This is called the diameter of S.

Theorem 4.9.21Let

{Fn }

n=1∞be a sequence of closed sets in F suchthat

nli→m∞ diam (Fn) = 0

and Fn⊇ Fn+1for each n. Then there exists a unique p ∈∩k=1∞Fk.

Proof: Pick pk∈ Fk. This is always possible because by assumption each set is nonempty.
Then

{pk}

k=m∞⊆ Fm and since the diameters converge to 0 it follows

{pk}

is a Cauchy
sequence. Therefore, it converges to a point, p by completeness of F. Since each Fk is closed,
it must be that p ∈ Fk for all k. Therefore, p ∈∩k=1∞Fk. If q ∈∩k=1∞Fk, then since both
p,q ∈ Fk,

|p− q| ≤ diam (Fk).

It follows since these diameters converge to 0,

|p− q|

≤ ε for every ε. Hence p = q.■

A sequence of sets,

{Gn}

which satisfies Gn⊇ Gn+1 for all n is called a nested sequence of
sets.

The next theorem is a major result called Bair’s theorem. It concerns the intersection of
dense open sets.

Definition 4.9.22An open set U ⊆ F is dense if for every x ∈ F andr > 0,B

(x,r)

∩ U≠∅.

Theorem 4.9.23Let

{U }
n

bea sequence of dense open sets. Then ∩nUnisdense.

Proof: Let p ∈ F and let r0> 0. I need to show D ∩B(p,r0)≠∅. Since U1 is dense, there
exists p1∈ U1∩ B(p,r0), an open set. Let p1∈ B(p1,r1) ⊆B(p1,r1)⊆ U1∩ B(p,r0) and
r1< 2−1. This is possible because U1∩ B

(p,r0)

is an open set and so there exists r1 such
that B

(p1,2r1)

⊆ U1∩ B

(p,r0)

. But

--------
B (p1,r1) ⊆ B (p1,r1) ⊆ B (p1,2r1)

because B

(p1,r1)

=

{x ∈ X : d(x,p) ≤ r1}

. (Why?)

PICT

There exists p2∈ U2∩ B(p1,r1) because U2 is dense. Let

p2 ∈ B (p2,r2) ⊆ B-(p2,r2) ⊆ U2 ∩B (p1,r1) ⊆ U1 ∩U2 ∩ B(p,r0).

and let r2< 2−2. Continue in this way. Thus

−n
rn < 2 ,

--------
B(pn,rn) ⊆ U1 ∩U2 ∩ ...∩Un ∩ B(p,r0),

--------
B(pn,rn) ⊆ B (pn−1,rn−1).

The sequence, {pn} is a Cauchy sequence because all terms of

{pk}

for k ≥ n are
contained in B

(pn,rn)

, a set whose diameter is no larger than 2−n. Since X is complete, there
exists p∞ such that

nli→m∞ pn = p∞.

Since all but finitely many terms of {pn} are in B(pm,rm), it follows that p∞∈B(pm,rm) for
each m. Therefore,